Question: $h(x) = x+6$ $f(x) = -4x^{2}-2x+2+3(h(x))$ $g(t) = -6t^{2}+2t+4(h(t))$ $ g(f(-2)) = {?} $
First, let's solve for the value of the inner function, $f(-2)$ . Then we'll know what to plug into the outer function. $f(-2) = -4(-2)^{2}+(-2)(-2)+2+3(h(-2))$ To solve for the value of $f$ , we need to solve for the value of $h(-2)$ $h(-2) = -2+6$ $h(-2) = 4$ That means $f(-2) = -4(-2)^{2}+(-2)(-2)+2+(3)(4)$ $f(-2) = 2$ Now we know that $f(-2) = 2$ . Let's solve for $g(f(-2))$ , which is $g(2)$ $g(2) = -6(2^{2})+(2)(2)+4(h(2))$ To solve for the value of $g$ , we need to solve for the value of $h(2)$ $h(2) = 2+6$ $h(2) = 8$ That means $g(2) = -6(2^{2})+(2)(2)+(4)(8)$ $g(2) = 12$